For which n are there primitive Pythagorean triples with legs of lengths a and a+n?

Question

For which n can $a^{2}+(a+n)^{2}=c^{2}$ be solved, where $a,b,c,n$ are positive integers? I have found solutions for $n=1,7,17,23,31,41,47,79,89$ and for multiples of $7,17,23$... Are there infinitely many prime $n$ for which it is solvable?

Answer 1

The general primite solution to $x^2+y^2 = z^2$ is given by: $x=u^2-v^2$, $y=2uv$, $z=u^2+v^2$, with $u,v$ relatively prime and not both odd.

For $(a,a+n,z)$ to be a primitive triple, we'd have to have a $(u,v)$ such that: $|u^2 - v^2 - 2uv| = n$. We can rewrite that as: $(u-v)^2 - 2v^2 = \pm n$

So, setting $w = u-v$, we want to find $(w,v)$ which are relatively prime and $w$ is odd, with:

$$w^2-2v^2 = \pm n$$

This means that $n$ must be odd.

In fact, we can use unique factorization in $\mathbb{Z}[\sqrt{2}]$ to show that $n$ can be any product of primes of the form $8k\pm 1$. Since there are infinitely many primes of the form $8k\pm 1$, the answer to your question is, "yes."

(Oh, and once you find one solution $(w,v)$ for a particular $n$, you can find infinitely many solutions for that $n$.)

Answer 2

$2a^{2}+2na+n^{2}=c^{2}$ --> $a=-\frac{-n+\sqrt{2c^{2}-n^{2}}}{2}$ --> there are solutions iff $x^{2}+n^{2}=2c^{2}$ has solutions --> find the set of the squares of all integers 0 in the set such that $y=2x-n$ then there is a primitive pythagorean triple with a difference of n between legs, and also for any multiple An if n>1 since if $k^{2}\equiv x(\mod{n})$ then $(Ak)^{2} \equiv Ax(\mod{An})$ --> $Ay=2Ax-An$.

Answer 3

If you solve expression for $n$ you get

$n=\sqrt{c^2-a^2}-a$, let's denote $b=\sqrt{c^2-a^2}$,so we have that $n=b-a$

Now,take look at picture bellow.Note that $AD=a$,and $BD=b-a=n$

If you change value of $b$ and keep $a$ to be constant you will get a infinite number of right triangles,and therefore infinite number of values of $n=b-a$,so answer is yes, there are infinitely many primes $n$ for which equation is solvable.